This is not really calculus. The question might have been answered more promptly, if asked in the algebra section. You also had a fair chance of an answer in the algebra . com website.
16x² +25y² -160x - 200y + 400 = 0
16x² -160x+25y² - 200y = - 400
16(x² -10x)+25(y² - 8y) = - 400
16(x² -10x)+16(25) +25(y² - 8y) +25(16)= - 400+16(25)+25(16)
16(x² -10x+25)+25(y² - 8y+16) = 400
16(x-5)^{2}+25(y-4)^{2} = 400
16(x-5)^{2}/400+25(y-4)^{2}/400 = 1
(x-5)^{2}/25+(y-4)^{2}/16 = 1
(x-5)^{2}/5^{2}+(y-4)^{2}/4^{2} = 1
You can see that the last equation above represents an ellipse
centered at (5,4),
with horizontal major axis,
a semi-major axis a=5 ,
and a semi-minor axis b=4 .
That means that the vertices, at the ends of the major axis, are
(5-5,4) =(0,4) and (5+5,4) = (10,4) .
The co-vertices, at the end of the minor axis, are
(5,4-4) = (5,0) and (5,4+4) = (5,8) .
The foci are on the major axis, between the center and the vertices,
at (5-c,4) and (5+c,4) .
All that is left to do is find the focal distance, c .
For that you may remember that in an ellipse a^{2} = b^{2} + c^{2} ,
or you may remember the definition of ellipse, and deduce the formula yourself.
In this case, substituting the values found for a and b,
5^{2} = 4^{2} + c^{2} , or
25 = 16 + c^{2} ---> c^{2} = 25-16 = 9 ---> c = 3 .
Then, the foci are at
(5-3,4) = (2,4) and (5+3,4) = (8,4) .

Sir, pwede din bang gamitin ang y' = infinity para kunin ang left and right vertices? In this way kasi, isa na lang ang kukunin, yung y' na lang, hindi na kailangan ang x'.

This is not really calculus. The question might have been answered more promptly, if asked in the algebra section. You also had a fair chance of an answer in the algebra . com website.

16x² +25y² -160x - 200y + 400 = 0

16x² -160x+25y² - 200y = - 400

16(x² -10x)+25(y² - 8y) = - 400

16(x² -10x)+16(25) +25(y² - 8y) +25(16)= - 400+16(25)+25(16)

16(x² -10x+25)+25(y² - 8y+16) = 400

16(x-5)

^{2}+25(y-4)^{2}= 40016(x-5)

^{2}/400+25(y-4)^{2}/400 = 1(x-5)

^{2}/25+(y-4)^{2}/16 = 1(x-5)

^{2}/5^{2}+(y-4)^{2}/4^{2}= 1You can see that the last equation above represents an ellipse

centered at (5,4),

with horizontal major axis,

a semi-major axis a=5 ,

and a semi-minor axis b=4 .

That means that the vertices, at the ends of the major axis, are

(5-5,4) =(0,4) and (5+5,4) = (10,4) .

The co-vertices, at the end of the minor axis, are

(5,4-4) = (5,0) and (5,4+4) = (5,8) .

The foci are on the major axis, between the center and the vertices,

at (5-c,4) and (5+c,4) .

All that is left to do is find the focal distance, c .

For that you may remember that in an ellipse a

^{2}= b^{2}+ c^{2},or you may remember the definition of ellipse, and deduce the formula yourself.

In this case, substituting the values found for a and b,

5

^{2}= 4^{2}+ c^{2}, or25 = 16 + c

^{2}---> c^{2}= 25-16 = 9 ---> c = 3 .Then, the foci are at

(5-3,4) = (2,4) and (5+3,4) = (8,4) .

Using Calculus for Vertices

$16x^2 + 25y^2 - 160x - 200y + 400 = 0$

For Upper and Lower Vertices

At the highest and lowest points, y' = 0

$32x - 160 = 0$

$x = 5$

$16(5^2) + 25y^2 - 160(5) - 200y + 400 = 0$

$25y^2 - 200y = 0$

$y = 8 ~ \text{and} ~ 0$

upper vertex = (5, 8)

lower vertex = (5, 0)

For Left and Right Vertices

At the extreme left and extreme right points, x' = 0

$50y - 200 = 0$

$y = 4$

$16x^2 + 25(4^2) - 160x - 200(4) + 400 = 0$

$16x^2 - 160x = 0$

$x = 10 ~ \text{and} ~ 0$

right vertex = (10, 4)

left vertex = (0, 4)

Below is the plot to locate the center and other points:

Sir, pwede din bang gamitin ang y' = infinity para kunin ang left and right vertices? In this way kasi, isa na lang ang kukunin, yung y' na lang, hindi na kailangan ang x'.